Differentiation Under the Integral Sign

Let $F(x) = \int_0^1 t^x\,dt$ for $x > 0$ . Show by direct computation that $F(x) = \frac{1}{x+1}$ and $F'(x) = -\frac{1}{(x+1)^2}$ , and prove completely, by direct substitution, that differentiating under the integral sign gives the same value: $\int_0^1 t^x \ln t\,dt = -\frac{1}{(x+1)^2}$ .

F(x) = \frac{1}{x+1}, \qquad F'(x) = -\frac{1}{(x+1)^2} = \int_0^1 t^x \ln t\,dt

x = 1

,

n = 200

: quadrature

\int_0^1 t\,dt = 0.5

(matches

\frac12

),

\int_0^1 t\ln t\,dt \approx -0.24999807

— within

2 \times 10^{-6}

of

-\frac14

.

Show proof

Fix $x > 0$ . On $t \in (0, 1]$ an antiderivative of $t^x$ is $\frac{t^{x+1}}{x+1}$ , and $t^{x+1} \to 0$ as $t \to 0^+$ since $x + 1 > 0$ . Hence $F(x) = \int_0^1 t^x\,dt = \left[\frac{t^{x+1}}{x+1}\right]_0^1 = \frac{1}{x+1}$ . Differentiating in $x$ gives $F'(x) = -\frac{1}{(x+1)^2}$ .
Now differentiate under the integral sign. For $t \in (0, 1]$ , $\frac{\partial}{\partial x} t^x = t^x \ln t$ (differentiate $t^x = e^{x \ln t}$ ). At $t = 0$ define the integrand as $0$ — because $\lim_{t \to 0^+} t^x \ln t = 0$ for $x > 0$ .
Compute $\int_0^1 t^x \ln t\,dt$ with $u = -(x+1)\ln t$ . For $t \in (0, 1]$ , $u \in [0, \infty)$ , $t = e^{-u/(x+1)}$ , $dt = -\frac{1}{x+1}e^{-u/(x+1)}\,du$ . Then $t^x \ln t\,dt = \frac{u e^{-u}}{(x+1)^2}\,du$ , so $\int_0^1 t^x \ln t\,dt = -\frac{1}{(x+1)^2}\int_0^\infty u e^{-u}\,du$ . By parts, $\int_0^\infty u e^{-u}\,du = \left[-u e^{-u}\right]_0^\infty + \int_0^\infty e^{-u}\,du = 0 + 1 = 1$ ; hence the under-integral derivative equals $-\frac{1}{(x+1)^2}$ — exactly the directly computed $F'(x)$ .
Numerically: at $x = 1$ the analytic values are $F(1) = \frac12$ , $F'(1) = -\frac14$ . In the scene with $n = 200$ Simpson panels, $\int_0^1 t\,dt = 0.5$ exactly, and $\int_0^1 t\ln t\,dt$ lands within $2 \times 10^{-6}$ of $-\frac14$ — all three estimates converge to one point.

Let $f(t, x) = t^x$ , $\frac{\partial f}{\partial x}(t, x) = t^x \ln t$ ( $t \in [0, 1]$ , defined as $0$ at $t = 0$ ). For $0 < \delta < A$ , show that $f$ and $\frac{\partial f}{\partial x}$ are continuous on the rectangle $[\delta, A] \times [0, 1]$ , and use this to prove rigorously that $F'(x) = \int_0^1 t^x \ln t\,dt$ for every $x > 0$ .

f, f_x \text{ continuous on } [\delta, A] \times [0, 1] \;\Rightarrow\; \frac{d}{dx}\int_0^1 f\,dt = \int_0^1 f_x\,dt

\left|t^x \ln t\right| \le t^{\delta}\,|\ln t| \xrightarrow[t \to 0^+]{} 0 \quad (x \ge \delta > 0)

x = 2

,

n = 200

: the quadrature of

\int_0^1 t^2 \ln t\,dt

is within

10^{-8}

of

-\frac19

— the agreement that continuity on

[1, 4] \times [0, 1]

guarantees.

Show proof

First, continuity. For $t > 0$ , $t^x = e^{x\ln t}$ is continuous in $(t, x)$ , so both $f$ and $f_x$ are continuous there. At $t = 0$ we defined the value as $0$ ; for $x \ge \delta > 0$ , $|t^x \ln t| = t^x |\ln t| \le t^{\delta}|\ln t| \to 0$ as $t \to 0^+$ , uniformly in $x \in [\delta, A]$ . Hence $f$ and $f_x$ are continuous on the compact rectangle $[\delta, A] \times [0, 1]$ .
Now prove the theorem: if $f$ , $f_x$ are continuous on $[a, b] \times [c, d]$ , then $F(x) = \int_c^d f(t, x)\,dt$ is differentiable with $F'(x_0) = \int_c^d f_x(t, x_0)\,dt$ . The difference quotient is $\frac{F(x_0+h) - F(x_0)}{h} = \int_c^d \frac{f(t, x_0+h) - f(t, x_0)}{h}\,dt$ . By the mean value theorem in $x$ , for each $t$ there is $\xi_{h,t}$ between $x_0$ and $x_0+h$ with $\frac{f(t, x_0+h) - f(t, x_0)}{h} = f_x(t, \xi_{h,t})$ . Since $f_x$ is uniformly continuous on the compact rectangle, $f_x(t, \xi_{h,t}) \to f_x(t, x_0)$ uniformly in $t$ as $h \to 0$ . Hence $\left|\frac{F(x_0+h)-F(x_0)}{h} - \int_c^d f_x(t, x_0)\,dt\right| \le (d-c)\sup_t |f_x(t, \xi_{h,t}) - f_x(t, x_0)| \to 0$ — the limit-integral interchange is justified.
For any $x_0 > 0$ , put $\delta = x_0/2$ , $A = 2x_0$ , so $x_0$ is interior to $[\delta, A]$ . Continuity from step 1 plus the theorem from step 2 give $F'(x_0) = \int_0^1 t^{x_0}\ln t\,dt$ , which equals $-\frac{1}{(x_0+1)^2}$ by the direct computation of problem 1. So the three estimates meeting in the scene is no accident — it is forced by uniform continuity on the compact rectangle.
Numerically: at $x = 2$ the analytic values are $F(2) = \frac13$ , $F'(2) = -\frac19$ . With $n = 200$ the Simpson approximation of $\int_0^1 t^2 \ln t\,dt$ is within $10^{-8}$ . Even though $\ln t$ blows up near $t = 0$ , the factor $t^2$ tames it ( $t^2|\ln t| \to 0$ ), and the numerics show the integral converging cleanly.

Let $\varphi(u) = 30u^2(1-u)^2$ ( $0 < u < 1$ ), $0$ elsewhere, and $f(x, t) = \varphi(t/x)$ ( $x > 0$ , $0 < t < x$ ), $0$ elsewhere ( $x, t \in [0, 1]$ ). With $F(x) = \int_0^1 f(x, t)\,dt$ , show (a) $F(x) = x$ for $0 < x < 1$ , so the right derivative at $0$ is $1$ ; (b) $\frac{\partial f}{\partial x}(0, t) = 0$ for every $t \in [0, 1]$ , so $\int_0^1 \frac{\partial f}{\partial x}(0, t)\,dt = 0$ ; (c) hence the under-integral differentiation formula fails at $x = 0$ . (d) Pinpoint where the hypotheses break.

F(x) = x\;(0 < x < 1), \quad F'_+(0) = 1 \ne 0 = \int_0^1 f_x(0, t)\,dt

f(x, x/2) = \tfrac{15}{8} \not\to 0 = f(0, 0), \qquad f_x(x, x/4) = -\tfrac{45}{32x} \to -\infty

This problem is about the "bump" counterexample, not the scene's

t^x

— the preset is the scene default

x = 1

,

n = 200

.

Show proof

First check $\int_0^1 \varphi = 1$ : $\int_0^1 30u^2(1-u)^2\,du = 30\int_0^1 (u^2 - 2u^3 + u^4)\,du = 30\left(\frac13 - \frac12 + \frac15\right) = 30 \cdot \frac{1}{30} = 1$ . For $0 < x < 1$ , $F(x) = \int_0^1 f(x, t)\,dt = \int_0^x 30(t/x)^2(1-t/x)^2\,dt$ (nonzero only for $0 < t < x$ ). With $u = t/x$ , $F(x) = x\int_0^1 30u^2(1-u)^2\,du = x \cdot 1 = x$ . Since $F(0) = 0$ , the right derivative at $0$ is $\lim_{h \to 0^+} \frac{F(h) - F(0)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$ .
Next, $\frac{\partial f}{\partial x}(0, t) = 0$ . Fix $t > 0$ : for $0 < h < t$ we have $t < h$ false, so $f(h, t) = 0$ ( $f$ is nonzero only when $0 < t < h$ ). Hence the difference quotient $\frac{f(h, t) - f(0, t)}{h} = 0$ for every $h < t$ , and the limit is $0$ . At $t = 0$ , $f(h, 0) = 0$ too. So $\frac{\partial f}{\partial x}(0, t) = 0$ for every $t \in [0, 1]$ , and $\int_0^1 \frac{\partial f}{\partial x}(0, t)\,dt = 0 \ne 1$ — different from the right derivative. The under-integral formula breaks at $x = 0$ .
Now locate the broken hypothesis. On the line $t = x/2$ , $f(x, x/2) = 30\left(\frac12\right)^2\left(\frac12\right)^2 = \frac{15}{8} = 1.875$ , independent of $x$ ; so $f \to 1.875 \ne 0 = f(0, 0)$ along $(x, x/2) \to (0, 0)$ — $f$ itself is discontinuous at $(0, 0)$ . The partial derivative is worse: with $u = t/x$ , $\frac{\partial f}{\partial x} = -\frac{60u^2(1-u)(1-2u)}{x}$ ( $0 < t < x$ ), and on the line $t = x/4$ ( $u = 1/4$ ) this is $-\frac{45}{32x} \to -\infty$ as $x \to 0^+$ . So $f_x$ is unbounded in every neighborhood of $(0, 0)$ , cannot be uniformly continuous, and the key step of problem 2's proof — uniform convergence of the difference quotients — collapses. Without the continuity hypotheses, the limit-integral interchange is not guaranteed — that is the moral of the counterexample.

Differentiation Under the Integral Sign

Examples & Exercises